Lec18 - Scott Hughes 22 April 2004 Massachusetts Institute...

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Unformatted text preview: Scott Hughes 22 April 2004 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Lecture 18: AC Circuits. Power and energy; Resonance. Filters. 18.1 AC circuits: Recap and summary Last time, we looked at AC circuits and found that they are quite simple to analyze provided we follows some simple rules: 1. Work with complex valued voltages and currents. Our driving AC EMF is usually some- thing like E ( t ) = E cos t ; replace this with E ( t ) = E e it (so that E ( t ) = Re[ E ( t )]). 2. The voltage drop across any circuit element obeys a generalized, complex version of Ohms law: V X = I X Z X where Z X is the impedance of circuit element X . Impedance works just like resistance, but is complex and frequency dependent: Z R = R Z C = 1 / ( iC ) Z L = iL . 3. Analyze the circuit as though it were a simple DC circuit containing only resistors as circuit elements. You may find phasor diagrams helpful for making sure that you get the magnitudes and phases correct. 4. Take the real part at the end of the day. 18.2 Power delivered to an AC circuit Lets look again at our prototypical driven RLC circuit: C L R Assume that the EMF supplied is E ( t ) = E cos t . In the complex representation, this becomes E ( t ) = E e it . We can now solve for the complex current in this circuit: E = V R + V L + V C = I R + iL + 1 iC - I = E R + i [ L- 1 / ( C )] . We now substitute E ( t ) = E e it , I ( t ) = I e- i e it . Then, we find I = E q R 2 + [ L- 1 / ( C )] 2 E / | Z tot ( ) | where | Z tot ( ) | is the magnitude of the total impedance. The phase is defined by tan =- Im[ I ] Re[ I ] = L R- 1 CR . The real current flowing in this circuit is thus given by I ( t ) = I cos( t- ) where I = E | Z tot ( ) | tan = L R- 1 CR . How much power is delivered to this circuit? From the definition of EMF as work per unit charge, and from the definition of current as charge per unit time, we know that the power delivered to this circuit must be P ( t ) = I ( t ) E ( t ) . Lets plug in the results we have for I ( t ) and E ( t ): P ( t ) = E 2 | Z tot ( ) | cos t cos( t- ) = E 2 | Z tot ( ) | [cos t cos t cos + cos t sin t sin ] . On the second line, I have used a trig identity, cos( a- b ) = cos a cos b +sin a sin b , to expand the second cosine....
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This note was uploaded on 04/11/2008 for the course PHYSICS 8.022 taught by Professor Hughes during the Spring '08 term at MIT.

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Lec18 - Scott Hughes 22 April 2004 Massachusetts Institute...

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